Jumat, 06 Mei 2011

Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

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Today we're going to continue with playing with liquids.

If I have an object that floats, a simple cylinder that floats in some liquid, the area is A here, the mass of the cylinder is M.

The density of the cylinder is rho and its length is l and the surface area is A.

So this is l.

And let the liquid line be here, and the fluid has a density rho fluid.

I call this level y1, this level y2.

The separation is h, and right on top here, there is the atmospheric pressure P2, which is the same as it is here on the liquid.

And here we have a pressure P1 in the liquid.

For this object to float we need equilibrium between, on the one hand, the force Mg and the buoyant force.

There is a force up here which I call F1, and there is a force down here which I call F2--

barometric pressure.

The force is always perpendicular to the surface.

There couldn't be any tangential component because then the air starts to flow, and it's static.

And here we have F1, which contains the hydrostatic pressure.

So P1 minus P2--

as we learned last time from Pascal--

equals rho of the fluid g to the minus y2 minus y1, which is h.

So that's the difference between the pressure P1 and P2.

For this to be in equilibrium, F1 minus F2 minus Mg has to be zero, and this we call the buoyant force.

And "buoyant" is spelt in a very strange way: b-u-o-y-a-n-t.

I always have to think about that.

It's the buoyant force.

F1 equals the area times P1 and F2 is the area times P2, so it is the area times P1 minus P2, and that is rho fluids times g times h.

And when you look at this, this is exactly the weight of the displaced fluid.

The area times h is the volume of the fluid which is displaced by this cylinder, and you multiply it by its density, that gives it mass.

Multiply it by g, that gives it weight.

So this is the weight of the displaced fluids.

And this is a very special case of a general principle which is called Archimedes' principle.

Archimedes' principle is as follows: The buoyant force on an immersed body has the same magnitude as the weight of the fluid which is displaced by the body.

According to legend Archimedes thought about this while he was taking a bath, and I have a picture of that here--

I don't know from when that dates--

but you see him there in his bath, but what you also see are there are two crowns.

And there is a reason why those crowns are there.

Archimedes lived in the third century B.C.

Archimedes had been given the task to determine whether a crown that was made for King Hieron II was pure gold.

The problem for him was to determine the density of this crown--

which is a very irregular-shaped object--

without destroying it.

And the legend has it that as Archimedes was taking a bath, he found the solution.

He rushed naked through the streets of Syracuse and he shouted, "Eureka! Eureka! Eureka!" which means, "I found it! I found it!" What did he find? What did he think of? He had the great vision to do the following: You take the crown and you weigh it in a normal way.

So the weight of the crown--

I call it W1--

is the volume of the crown times the density of which it is made.

If it is gold, it should be 19.3, I believe, and so this is the mass of the crown and this is the weight of the crown.

Now he takes the crown and he immerses it in water.

And he has a spring balance, and he weighs it again.

And he finds that the weight is less and so now we have the weight immersed in water.

So what you get is the weight of the crown minus the buoyant force, which is the weight of the displaced fluid.

And the weight of the displaced fluid is the volume of the crown--

because the crown is where...

the water has been removed where the crown is--

times the density of the fluid--

which is water, which he knew very well--

times g.

And so this part here is weight loss.

That's the loss of weight.

You can see that, you can measure that with a spring.

It's lost weight, because of the buoyant force.

And so now what he does, he takes W1 and divides that by the weight loss and that gives you this term divided by this term, which immediately gives you rho of the crown divided by rho of the water.

And he knows rho of the water, so he can find rho of the crown.

It's an amazing idea; he was a genius.

I don't know how the story ended, whether it was gold or not.

It probably was, because chances are that if it hadn't been gold that the king would have killed him--

for no good reason, but that's the way these things worked in those days.

This method is also used to measure the percentage of fat in persons' bodies, so they immerse them in water and then they weigh them and they compare that with their regular weight.

Let's look at an iceberg.

Here is an iceberg.

Here is the water--

it's floating in water.

It has mass M, it has a total volume V total, and the density of the ice is rho ice, which is 0.92 in grams per cubic centimeter.

It's less than water.

This is floating, and so there's equilibrium between Mg and the buoyant force.

So Mg must be equal to the buoyant force.

Now, Mg is the total volume times rho ice times g, just like the crown.

The buoyant force is the volume underwater, which is this part, times the density of water, rho water, times g.

You lose your g, and so you find that the volume underwater divided by the total volume equals rho ice divided by rho of water, which is 0.92.

That means 92% of the iceberg is underwater, and this explains something about the tragedy on April 15, 1912, when the Titanic hit an iceberg.

When you encounter an iceberg, you literally only see the tip of the iceberg.

That's where the expression comes from.

92% is underwater.

I want to return now to my cylinder, and I want to ask myself the question, when does that cylinder float? What is the condition for floating? Well, clearly, for that cylinder to float the buoyant force must be Mg, and the buoyant force is the area times h--

that's the volume underwater--

multiplied by the density of the fluid times g must be the total volume of the cylinder, which is the area times l, because that was the length of the cylinder, times the density of the object itself times g.

I lose my A, I lose my g, but I know that h must be less than l; otherwise it wouldn't be floating, right? The part below the water has to be smaller than the length of the cylinder.

And if h is less than l, that means that the density of the fluid must be larger than the density of the object, and this is a necessary condition for floating.

And therefore, if an object sinks then the density of the object is larger than the density of the fluid.

And the amazing thing is that this is completely independent of the dimensions of the object.

The only thing that matters is the density.

If you take a pebble and you throw it in the water, it sinks, because the density of a pebble is higher than water.

If you take a piece of wood, which has a density lower than water, and you throw it on water, it floats independent of its shape.

Whether it sinks or whether it floats, the buoyant force is always identical to the weight of the displaced fluid.

And this brings up one of my favorite questions that I have for you that I want you to think about.

And if you have a full understanding now of Archimedes' principle, you will be able to answer it, so concentrate on what I am going to present you with.

I am in a swimming pool, and I'm in a boat.

Here is the swimming pool and here is the boat, and I am sitting in the boat and I have a rock here in my boat.

I'm sitting in the swimming pool, nice rock in my boat.

I mark the waterline of the swimming pool very carefully.

I take the rock and I throw it overboard.

Will the waterline go up, or will the waterline go down, or maybe the waterline will stay the same? Now, use your intuition--

don't mind being wrong.

At home you have some time to think about it, and I am sure you will come up with the right answer.

Who thinks that the waterline will go up the swimming pool? Who thinks that the waterline will go down? Who thinks that it will make no difference, that the waterline stays the same?



Well, the waterline will change, but you figure it out.

Okay, you apply Archimedes' principle and you'll get the answer.

I want to talk about stability, particularly stability of ships, which is a very important thing--

they float.

Suppose I have an object here which is floating in water.

Here is the waterline, and let here be the center of mass of that object.

Could be way off center.

It could be an iceberg, it could be boulders, it could be rocks in there, right? It doesn't have to be uniform density.

The center of mass could be off the center...

of the geometric center.

So if this object has a certain mass, then this is the gravitational force.

But now look at the center of mass fluid that is displaced.

That's clearly more here, somewhere here, the displaced fluid.

That is where the buoyant force acts.

And so now what you have...

You have a torque on this object relative to any point that you choose.

It doesn't matter where you pick a point, you have a torque.

And so what's going to happen, this object is clearly going to rotate in this direction.

And the torque will only be zero when the buoyant force and the gravitational force are on one line.

Then the torque becomes zero, and then it is completely happy.

Now, there are two ways that you can get them on one line.

We discussed that earlier in a different context.

You can either have the center of mass of the object below the center of mass of the displaced fluid or above.

In both cases would they be on one line.

However, in one case, there would be stable equilibrium.

In the other, there would not be a stable equilibrium.

I have here an object which has its center of mass very low.

You can't tell that--

no way of knowing.

All you know is that the weight of the displaced fluid that you see here is the same as the weight of the object.

That's all you know.

If I took this object and I tilt it a little with the center of mass very low--

so here is Mg and here is somewhere the waterline--

so the center of mass of the displaced fluid is somewhere here, so Fb is here, the buoyant force, you can see what's going to happen.

It's going to rotate towards the right--

it's a restoring torque, and so it's completely stable.

I can wobble it back and forth and it is stable.

If I would turn it over, then it's not stable, because now I would have the center of mass somewhere here, high up, so now I have Mg.

And the center of the buoyant force, the displaced water, is about here, so now I have the buoyant force up, and now you see what's going to happen.

I tilt it to the side, and it will rotate even further.

This torque will drive it away from the vertical.

And that's very important, therefore, with ships, that you always build the ship such that the center of mass of the ship is as low as you can get it.

That gives you the most stable configuration.

If you bring the center of mass of ships very high--

in the 17th century, they had these very massive cannons which were very high on the deck--

then the ship can capsize, and it has happened many times because the center of mass was just too high.

So here... the center of mass is somewhere here.

Very heavy, this part.

And so now, if I lower it in the water notice it goes into the water to the same depth, because the buoyant force is, of course, the same, so the amount of displaced water is the same in both cases.

But now the center of mass is high and this is very unstable.

When I let it go, it flips over.

So the center of mass of the object was higher than the center of mass of the displaced fluid.

And so with ships, you have to be very careful about that.

Let's talk a little bit about balloons.

If I have a balloon, the situation is not too dissimilar from having an object floating in a liquid.

Let the balloon have a mass M.

That is the mass of the gas in the balloon plus all the rest, and what I mean by "all the rest"...

That is the material of the balloon and the string--

everything else that makes up the mass.

It has a certain volume v, and so there is a certain rho of the gas inside and there is rho of air outside.

And I want to evaluate what the criterion is for this balloon to rise.

Well, for it to rise, the buoyant force will have to be larger than Mg.

What is the buoyant force? That is the weight of the displaced fluid.

The fluid, in this case, is air.

So the weight of the displaced fluid is the volume times the density of the air--

that's the fluid in which it is now--

times g, that is the buoyant force.

That's... the weight of the displaced fluid has to be larger than Mg.

Now, Mg is the mass of the gas, which is the volume of the gas times the density of the gas.

That's the mass times g--

because we have to convert it to a force--

plus all the rest, times g.

I lose my g, and what you see...

that this, of course, is always larger than zero.

There's always some mass associated with the skin and in this case with the string.

But you see, the only way that this balloon can rise is that the density of the gas must be smaller than the density of air.

Density of the gas must be less than the density of the air.

This is a necessary condition for this to hold.

It is not a sufficient condition, because I can take a balloon, put a little bit of helium in there--

so the density of the gas is lower than the density of air--

but it may not rise, and that's because of this term.

But it is a necessary condition but not a sufficient condition.

Now I'm going to make you see a demonstration which is extremely nonintuitive, and I will try, step by step, to explain to you why you see what you see.

What you're going to see, very nonintuitive, so try to follow closely why you see what you will see.

I have here a pendulum with an apple, and here I have a balloon filled with helium.

I cut this string and I cut this string.

Gravity is in this direction.

The apple will fall, the balloon will rise.

The balloon goes in the opposite direction than the gravitational acceleration.

If there were no gravity, this balloon would not rise and the apple would not fall.

Do we agree so far? Without gravity, apple would not fall, balloon would not rise.

Now we go in outer space.

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

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